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LemmaStatement: Literature-sourcedProof: AI-adaptedverified 2026-09-26 (gpt-6-sol)
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For 1<p<∞, the Lp norm of a nonnegative function is the supremum of its pairings with Lq unit vectors

Statement

Let 1<p<∞ and let q be the conjugate exponent. If F∈Lp(μ) is nonnegative, then

∥F∥p=sup⁡{∫Fg dμ:g≥0, ∥g∥q≤1}.

Facts & Assumptions

Given: 1<p<∞ and a nonnegative function F∈Lp(μ).

[L1]
[L2]

Holder's inequality holds for the pairing (Holder's inequality for integrals, including the endpoint cases).

Proof

technique · direct
1.1L2L3givenalgebra

For every g≥0 with ∥g∥q≤1, [L2] gives ∫Fg dμ≤∥F∥p∥g∥q≤∥F∥p. So the displayed supremum is at most ∥F∥p.

2.1L3step 1.1construct

If ∥F∥p=0, then F=0 almost everywhere. Step 1.1 bounds the supremum above by 0, while the admissible choice g=0 gives value 0, so the supremum is 0.

2.2L1L3step 1.1algebraconstruct

If ∥F∥p>0, define g:=Fp−1∥F∥pp/q. Because (p−1)q=p, one has ∥g∥qq=∫F(p−1)q dμ∥F∥pp=1, so ∥g∥q=1, and ∫Fg dμ=∫Fp dμ∥F∥pp/q=∥F∥p.

3.1step 1.1step 2.1step 2.2∎

Step 1.1 gives the upper bound and steps 2.1–2.2 attain it in the two cases, so the supremum equals ∥F∥p.

Depends on

Used by

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Sources